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I'm not really sure how to go about creating the equation, can anyone help me?

A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the x axis be parallel to the ground and the positive y axis point upward. In the time interval from t=0 to t=4 s, the trajectory of the car along a certain section of the track is given by r(arrow on top) =A(1m/s)ti^+A[(1m/s3)t3−6(1m/s2)t2]j^, where A is a positive dimensionless constant.

Part B: Derive a general expression for the speed v of the car. Make sure that your expression would give the correct value for the speed (in m/s), but you don't need to put in anything explicitly about units (unlike what you see in the original expression for r(arrow on top)

Respuesta :

The displacement vector (SI units) is
[tex]\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}[/tex]

The speed is a scalar quantity. Its magnitude is
[tex]v= \sqrt{A^{2}t^{2}+A^{2}(t^{3}-6t^{2})^{2}} \\ v=A \sqrt{t^{2}+t^{6}-12t^{5}+36t^{4}} \\ v=At \sqrt{t^{4}-12t^{3}+36t^{2}+1} [/tex]

Answer: At√(t⁴ - 12t³ + 36t² + 1)
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